Mathmatical Induction Problem (Divisibility)

In summary, the conversation discusses using mathematical induction to prove that the expression 12^n + 2(5^{n-1}) is divisible by 7 for all positive integers n. The first step is to show that it works for n=1 and then assume it works for k. The attempt at a solution involves substituting values and rearranging the expression, but the correct step is to factor out 12^k and use the assumption that 12^k + 2(5^{k-1}) is divisible by 7.
  • #1
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Homework Statement


Use Mathematical Induction to prove that [tex] 12^n + 2(5^{n-1}) [/tex] is divisible by 7 for all [tex] n \in Z^+ [/tex]

Homework Equations


The Attempt at a Solution



First, show that it works for n = 1:
[tex] 12^1 + 2 \cdot 5^0 = 14 [/tex] , 14/7 = 2

Next assume:
[tex] 12^k + 2(5^{k-1}) = 7A [/tex]

Then, prove for k + 1:
[tex] 12^{k+1} + 2(5^k) [/tex]

I can't figure out how to prove this. I know that this can be changed to:
[tex] 12 \cdot 12^{k} + 2 \cdot 5 (5^{k-1}) [/tex]
But that doesn't seem to help me much.

I also tried substituting values for 12^k and 5^(k-1) from above:
[tex] 12^k = 7A - 2(5^{k-1}) [/tex]
[tex] 2(5^{k-1}) = 7A - 12^k [/tex]

This doesn't seem too help either, I can reduce it to:
[tex] 189A - (12 \cdot 2(5^{k-1})+5(12^k)) [/tex]

Any suggestions?
Thanks,
Tom
 
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  • #2
Actually, having come till

[tex]12.12^k + 2.5(5^{k-1})[/tex],

the next step should have been

[tex]7.12^k + 5.12^k + 2.5(5^{k-1})[/tex].
 
  • #3
Ah, got it now. Thank you. I don't like these induction problems...
 

1. What is mathematical induction?

Mathematical induction is a proof technique used to show that a statement or property holds for all natural numbers. It is based on the principle that if a statement is true for the first natural number, and if it is true for any given natural number, then it must also be true for the next natural number.

2. How is mathematical induction used to prove divisibility?

In order to prove a statement about divisibility using mathematical induction, we first show that the statement is true for the first natural number (usually 1). Next, we assume that the statement is true for some arbitrary natural number n. Finally, we use this assumption to prove that the statement is also true for the next natural number (n+1). This shows that the statement must be true for all natural numbers, including n+1, and therefore it holds for all natural numbers.

3. What is the difference between weak and strong mathematical induction?

Weak mathematical induction, also known as the principle of mathematical induction, uses the assumption that the statement being proved is true for an arbitrary natural number n to prove that it is true for the next natural number (n+1). Strong mathematical induction, on the other hand, uses the assumption that the statement is true for all natural numbers up to n to prove that it is true for n+1. Both methods are valid and can be used to prove statements about divisibility.

4. Can mathematical induction be used to prove non-divisibility?

Yes, mathematical induction can be used to prove statements about non-divisibility. For example, if we want to prove that a number cannot be divided by 3, we can use induction to show that for any natural number n, n+1 cannot be divided by 3 if n is not divisible by 3. This would then prove that n+1 is also not divisible by 3, and therefore the statement holds for all natural numbers.

5. What are common mistakes to avoid when using mathematical induction to prove divisibility?

One common mistake is assuming that the statement being proved is true for all natural numbers, without first showing that it is true for the first natural number (usually 1). Another mistake is assuming that the statement is true for an arbitrary natural number n, without proving that it is also true for n+1. It is important to clearly state the base case and the inductive step in order to have a valid proof by mathematical induction.

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